Answer:
a. y(t) = - 1/2gt² + y₀ b. ![t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} + 2 y_{0}g } }{g }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bv_%7B0%7D%20%2B%2F-%20%5Csqrt%7Bv_%7B0%7D%20%5E%7B2%7D%20%2B%202%20y_%7B0%7Dg%20%20%7D%20%20%20%7D%7Bg%20%7D)
Step-by-step explanation:
Let the quadratic function y(t) = v₀t - 1/2gt² + y₀ represent the quadratic function that models the height above the ground of the projectile.
a. Maximum Height
At maximum height, the velocity, v₀ = 0, so substituting v₀ = 0 into the equation, we have
y(t) = v₀t - 1/2gt² + y₀
y(t) = 0 × t - 1/2gt² + y₀
y(t) = 0 - 1/2gt² + y₀
y(t) = - 1/2gt² + y₀
b. Time when the projectile reaches the ground
The time when the projectile reaches the ground is gotten when y(t) = 0, So
y(t) = v₀t - 1/2gt² + y₀
0 = v₀t - 1/2gt² + y₀
re-arranging, we have
1/2gt² - v₀t - y₀ = 0
Using the quadratic formula,
![t = \frac{-(-v_{0}) +/- \sqrt{(-v_{0}) ^{2} - 4 X (-y_{0}) X\frac{g}{2} } }{2 X \frac{g}{2} } \\= \frac{v_{0} +/- \sqrt{v_{0} ^{2} + 2 y_{0}g } }{g }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-%28-v_%7B0%7D%29%20%2B%2F-%20%5Csqrt%7B%28-v_%7B0%7D%29%20%5E%7B2%7D%20-%204%20X%20%28-y_%7B0%7D%29%20X%5Cfrac%7Bg%7D%7B2%7D%20%20%7D%20%20%20%7D%7B2%20X%20%5Cfrac%7Bg%7D%7B2%7D%20%7D%20%5C%5C%3D%20%5Cfrac%7Bv_%7B0%7D%20%2B%2F-%20%5Csqrt%7Bv_%7B0%7D%20%5E%7B2%7D%20%2B%202%20y_%7B0%7Dg%20%20%7D%20%20%20%7D%7Bg%20%7D)